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To everyone: don't bother writing up another answer, i'm giving this bounty Antonio's answer. It just doesn't let me yet (24 hours delay).

If you map the nth roots of unity $z$ with the function $-W(-z/e)$ you get decent starting points for some root finding algorithm to the roots of the scaled truncated taylor series of $\exp$. Here W is the lambertW function, $e$ is $\exp(1)$ and 'scaled' in 'scaled truncated taylor series of exp' means the following: say if $$s_5(x) = 1+x+x^2/2+x^3/6+x^4/24+x^5/120$$ is the 'truncated taylor series of exp' of degree 5 then we will look at $s_5(5x)$ so we are looking at $s_n(nx)$ in general.

Here is a plot curve for the case $n=33$ (it only works for uneven $n$). This is formula (1.1) from paper 221 available from here - $-W(-z/e)$ is the inverse of $ze^{1-z}$.

formula 1.1

How to get a better map from the roots of unity to the roots of this polynomial? Alternatively, is there some infinite sum representation for the roots? There isn't much difference: "applying" LambertW to some start values is pretty much the same as an infinite series.

Here is an octave script for such a plot as the one above (To use lambertw(), as in the script, install the 'specfun' package for octave - or use a more number/function theory centric system than octave).

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up vote 5 down vote accepted

You may be interested in the paper Locating the zeros of partial sums of $\exp(z)$ with Riemann-Hilbert Methods by T. Kriecherbauer, A.B.J. Kuijlaars, K.D.T-R McLaughlin, and P.D. Miller (arXiv preprint available here). In section 4 they give asymptotic series for the zeros in terms of the images of the roots of unity through the map $z \mapsto -W(-z/e)$.

I'm not familiar with their methods, but I do know of another way to find asymptotic approximations for the zeros of $s_n(nz)$ which stay away from the point $z=1$ (that is, which remain in a compact subset of the punctured plane $\mathbb{C} \setminus \{1\}$ as $n \to \infty$).

The zeros of $s_n(nz)$ satisfy the asymptotic equation

$$ \left(ze^{1-z}\right)^n = \sqrt{2\pi n} \frac{1-z}{z} \Bigl(1+\epsilon_n(z)\Bigr), \tag{1} $$

where $\epsilon_n(z) = O(1/n)$ as long as $z$ remains in a compact subset of $\operatorname{Re}(z) < 1$ (at least). By solving this equation for $z$ one may find asymptotic expressions for the individual zeros.

For instance, when $n$ is odd, $s_n(nz)$ has a single real zero $z_n$ which approaches

$$ z=-W(1/e) \approx -0.278465 $$

as $n \to \infty$. For convenience let's define

$$ w = W(1/e). $$

According to the paper On the Zeroes of the Nth Partial Sum of the Exponential Series by S. Zemyan (JSTOR link), Szegő showed that

$$ z_n = -w - \frac{w}{(1+w)n} \log\left(\sqrt{2\pi n} \frac{1+w}{w}\right) + o\left(\frac{1}{n}\right) \tag{2} $$

as $n \to \infty$.

For this result Zemyan cites a book by Pólya and Szegő published in the 60s, though I'm sure Szegő wrote down something like this when he was originally investigating the zeros of these partial sums in the 20s.

In attempting to derive this result myself from equation $(1)$ I found the formula

$$ z_n = -w - \frac{w}{(1+w)n} \log\left(\sqrt{2\pi n} \frac{1+w}{w}\right) - \frac{w}{2(1+w)^3n^2} \left\{\frac{(\log n)^2}{4} + \left[\log\left(\sqrt{2\pi} \frac{1+w}{w}\right)-1\right]\log n\right\} + O\left(\frac{1}{n^2}\right), \tag{3} $$

which is a slight improvement on Szegő's approximation $(2)$. The calculation was tedious, to say the least, but the method can be generalized to find approximations for every such zero of $s_n(nz)$. Begin by writing $z = -W(-\zeta/e) + \delta$, where $\zeta$ is an $n^\text{th}$ root of $-1$, and solve $(1)$ for $\delta$ under the assumption that $\delta$ is small. (Note that in my calculation I chose $\zeta = -1$.)

In a sense this method was used in the paper Asymptotics for the zeros of the partial sums of $e^z$. I by A.J. Carpenter, R.S. Varga, and J. Waldvogel (Project Euclid link), though they didn't carry it through as such. I believe it was actually used prior to that in Carpenter's doctoral thesis.

Below is a plot of the numerical solutions to $s_{2n+1}((2n+1)z) = 0$ near $z=-W(1/e)$ as black dots, Szegő's approximation $(2)$ as a blue line, and the approximation in $(3)$ as a red line for $20 \leq n \leq 40$.

enter image description here

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Theorem 4.1 and 4.2 answer the question of an infinite series expansions for the roots (am i reading that right?). Are the $\alpha_{k,n}$ given by $-W(-\zeta/e)$? Followup question: how does the series for $Q_i(x,y)$ from theorem 1 and $q_i(x)$ from theorem 2 continue? Have they published the follow up or is there some sort of announcement regarding publication? The equations at the beginning of section 4 look a lot like the definition of LambertW. Where do the roots of unity come in? – Peter Sheldrick Nov 5 '13 at 21:05
Also, thank you very much for your work. – Peter Sheldrick Nov 5 '13 at 21:09
You're very welcome :). And yes, $$\alpha_{k,n} = -W\left(-\frac{e^{2\pi i k/n}}{e}\right),$$ if I'm reading it correctly. I believe they're taking the principal branch of $W$. – Antonio Vargas Nov 5 '13 at 21:11
As for the polynomials $Q_i$ and $q_i$, I'm not sure if there's a nice way to calculate them. I suppose they might satisfy some kind of recurrence relation, but showing that might take some work. And yes, $A_n$ may be chosen arbitrarily; they indicate that better approximations can be achieved by choosing it more appropriately but at the cost of the nice representation of the roots $\alpha_{k,n}$, for instance. – Antonio Vargas Nov 5 '13 at 21:37
@PeterSheldrick You accepted the answer but did not award the bounty. Why so? – Ahaan S. Rungta Nov 6 '13 at 16:27

The answers to question 4329 over on Math Overflow might be helpful to you.

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yeah thanks - i saw that one as well. It was definitly one of the most helpful links i found (and you also found). They talk about the 'szegö-curve' $|se^{1-s}|=1$ that the roots of the scaled taylor polynomials converge to. But what about the roots for fixed $n$ as opposed to $n\rightarrow $inf?. Why do the roots of unity under $-W(-z/e)$ fit so well for the roots of the taylor polynomials? – Peter Sheldrick Jul 15 '11 at 10:20

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